How do you integrate int x^2e^(x^3/2)dx from [-2,3]?

1 Answer
Jan 14, 2017

(2e^(35/2)-2)/(3e^4)

Explanation:

I=int_(-2)^3x^2e^(x^3/2)dx

Use the substitution:

u=x^3/2" "=>" "du=3/2x^2dx

We already have x^2dx in the integral, but are off by a factor of 3/2. Also note that changing from x to u will change our bounds:

x=3" "=>" "u=3^3/2=27/2

x=-2" "=>" "u=(-2)^3/2=-4

Thus:

I=2/3inte^(x^3/2)(3/2x^2dx)=2/3int_(-4)^(27/2)e^udu

The integral of e^u is itself:

=2/3[e^u]_(-4)^(27/2)=2/3(e^(27/2)-e^(-4))=2/3((e^(35/2)-1)/e^4)=(2e^(35/2)-2)/(3e^4)